Problem 19 - Chemistry The Central Science

Question

The radius of an atom of gold (Au) is about 1.35 \(\mathrm{A}\) .(a) Express this distance in nanometers \((\mathrm{nm})\) and in picometers \((\mathrm{pm}) .\) (b) How many gold atoms would have to be lined up to span 1.0 \(\mathrm{mm}\) ? (c) If the atom is assumed to be a sphere,what is the volume in \(\mathrm{cm}^{3}\) of a single Au atom?

Step-by-Step Solution

Answer

(a) The radius of a gold atom in nanometers and picometers is 0.135 nm and 135 pm. (b) Approximately 370,370 gold atoms would have to be lined up to span 1.0 mm. (c) The volume of a single gold atom, assuming it's a sphere, is approximately 1.03 * 10⁻²³ cm³.

Step 1: (a) Conversion to nanometers and picometers

First, we need to convert the radius of the gold atom from Angstroms to nanometers and picometers. The conversion factors are: - 1 Angstrom (A) = 0.1 nanometers (nm) - 1 Angstrom (A) = 100 picometers (pm) So, to convert: 1.35 A * 0.1 nm/A = 0.135 nm (nanometers) 1.35 A * 100 pm/A = 135 pm (picometers)

Step 2: (b) Number of gold atoms to span 1.0 mm

To find out how many gold atoms are needed to span 1.0 mm, we can first consider the diameter of each gold atom, which is 2 times the given radius. Diameter of 1 gold atom = 2 * 1.35 A = 2.7 A Now, we need to convert 1.0 mm to Angstroms. The conversion factor is: - 1 millimeter (mm) = 10³ µm (micrometers) = 10³ * 10² nm (nanometers) = 10₃ * 10² * 10 A (Angstroms) 1.0 mm in Angstroms: 1.0 mm * 10³ * 10² * 10 A/mm = 10⁶ A Next, we can divide the total distance in Angstroms by the diameter of 1 gold atom to find the number of atoms required: Number of atoms = (10⁶ A) / (2.7 A) ≈ 370,370 gold atoms

Step 3: (c) Volume of a single gold atom

To calculate the volume of a single gold atom, we need to use the formula for the volume of a sphere: Volume = \(\frac{4}{3}\pi r^{3}\) Since the radius of the gold atom is given in Angstroms, we need to convert it to centimeters first. The conversion factor is: - 1 Angstrom (A) = 10⁻⁸ cm Converting the radius to centimeters: 1.35 A * 10⁻⁸ cm/A = 1.35 * 10⁻⁸ cm Now, we can plug the radius in centimeters into the formula for the volume of a sphere: Volume = \(\frac{4}{3}\pi (1.35 * 10⁻⁸)^{3}\) ≈ 1.03 * 10⁻²³ cm³ So, the volume of a single gold atom (considered as a sphere) is approximately 1.03 * 10⁻²³ cm³.

Key Concepts

Angstrom to Nanometer Conversion

Understanding the conversion of angstroms to nanometers is essential when studying atomic dimensions, which are extremely small. An angstrom, denoted as \( \AA \), is a unit of length used to express atomic and molecular dimensions. One angstrom is equivalent to 0.1 nanometers (nm), or \(1 \times 10^{-10} \text{meters}\). To convert angstroms to nanometers, simply multiply the number of angstroms by 0.1. For example, if the radius of a gold atom is 1.35 angstroms, to convert it to nanometers, you would perform the following calculation:

\[ 1.35 \; \AA \times 0.1 \; \frac{\text{nm}}{\AA} = 0.135 \; \text{nm} \]
Converted to nanometers, the radius of a gold atom would be 0.135 nm. This conversion is commonly used in chemistry and physics to describe the scale at which atomic and molecular interactions occur.

Angstrom to Picometer Conversion

Similarly, understanding the conversion from angstroms to picometers is another critical aspect of dealing with atomic scale measurements. Since a picometer is one trillionth of a meter, or \( 1 \times 10^{-12} \text{meters}\), it is frequently used to represent the minutely small distances between atoms in crystals or molecules. One angstrom is equal to 100 picometers (pm). To convert angstroms to picometers, you would multiply the number of angstroms by 100.

Taking the gold atom radius from the earlier example, conversion to picometers would be:

\[ 1.35 \; \AA \times 100 \; \frac{\text{pm}}{\AA} = 135 \; \text{pm} \]
After the conversion, the radius of the gold atom would then be 135 pm. This unit is especially relevant in fields such as crystallography and nanotechnology, where distances smaller than a nanometer are significant.

Volume of an Atom

When the concept of atomic radius is understood, estimating the volume of an atom becomes an interesting exercise in applying mathematical formulas to real-world concepts. To determine the volume of an atom, which is typically approximated as a sphere, we use the formula for the volume of a sphere:

\[ \text{Volume} = \frac{4}{3}\pi r^{3} \]
First, the radius should be in consistent units suitable for volume calculation. As given in the example of a gold atom, the radius in angstrom translates to centimeters using the conversion \(1 \AA = 10^{-8} \text{cm}\). After conversion, the radius is then cubed, and the result is multiplied by \(\pi\) and then by \(\frac{4}{3}\) to obtain the volume in cubic centimeters (\(\text{cm}^3\)). For instance, the volume of a gold atom with a radius of 1.35 angstroms is calculated as follows:

\[ \left(1.35 \times 10^{-8}\right)^{3} \text{cm}^{3} \times \frac{4}{3}\pi \approx 1.03 \times 10^{-23} \text{cm}^{3} \]
Thus, one can visually conceptualize the space a gold atom occupies in three dimensions, despite being imperceptibly small. Understanding atomic volume is beneficial in many areas of science, including material science and chemistry, where the physical arrangement of atoms determines a material's properties.